The $n^{\text{th}}$ derivative of $g$ at $x=0$ is given by $g^{(n)}(0)=\dfrac{\sqrt{n+7}}{n^3}$ for $n\ge 1$. What is the coefficient for the term containing $x^2$ in the Maclaurin series of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{16}$ (Choice B) B $\dfrac{3}{64}$ (Choice C) C $\dfrac{3}{8}$ (Choice D) D $\dfrac{3}{32}$
Explanation: The quadratic term of the Taylor series centered at $~x=0~$ is $~{g}\,^{\prime\prime}\left( 0 \right)\frac{{{x}^{2}}}{2!}\,$. From the given information, $~{g}\,^{\prime\prime}\left( 0 \right)=\frac{\sqrt{9}}{8}=\frac{3}{8}\,$. Then the coefficient of $~{{x}^{2}}~$ is $~\frac{3}{8}\cdot \frac{1}{2!}=\frac{3}{16}\,$.